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f-t-0-t-x-x-dx-




Question Number 180786 by Vynho last updated on 17/Nov/22
f(t)=∫_0 ^t x−⌊x⌋  dx
$${f}\left({t}\right)=\int_{\mathrm{0}} ^{{t}} {x}−\lfloor{x}\rfloor\:\:{dx} \\ $$
Answered by mr W last updated on 17/Nov/22
let n=⌊t⌋  f(t)=Σ_(k=0) ^(n−1) ∫_k ^(k+1) (x−⌊x⌋)dx+∫_n ^t (x−⌊x⌋)dx  f(t)=Σ_(k=0) ^(n−1) ∫_0 ^1 (k+ξ−k)d(k+ξ)+∫_0 ^(t−n) (n+ξ−n)d(n+ξ)  f(t)=Σ_(k=0) ^(n−1) ∫_0 ^1 ξdξ+∫_0 ^(t−n) ξdξ  f(t)=Σ_(k=0) ^(n−1) ((1/2))+(((t−n)^2 )/2)  f(t)=(n/2)+(((t−n)^2 )/2)  f(t)=((⌊t⌋+(t−⌊t⌋)^2 )/2)
$${let}\:{n}=\lfloor{t}\rfloor \\ $$$${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \left({x}−\lfloor{x}\rfloor\right){dx}+\int_{{n}} ^{{t}} \left({x}−\lfloor{x}\rfloor\right){dx} \\ $$$${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({k}+\xi−{k}\right){d}\left({k}+\xi\right)+\int_{\mathrm{0}} ^{{t}−{n}} \left({n}+\xi−{n}\right){d}\left({n}+\xi\right) \\ $$$${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \xi{d}\xi+\int_{\mathrm{0}} ^{{t}−{n}} \xi{d}\xi \\ $$$${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\left({t}−{n}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${f}\left({t}\right)=\frac{{n}}{\mathrm{2}}+\frac{\left({t}−{n}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${f}\left({t}\right)=\frac{\lfloor{t}\rfloor+\left({t}−\lfloor{t}\rfloor\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$

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